SOLUTION: what are the 2 x-intercept coordinates when the vertex is (-2,-8) and the y-intercept is (0,4)?

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Question 739832: what are the 2 x-intercept coordinates when the vertex is (-2,-8) and the y-intercept is (0,4)?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The general form is:
+y+=+ax%5E2+%2B+b%2Ax+%2B+c+
The vertex is at ( -2,-8 )
The formula for the x-intercept of the vertex is:
+x%5Bv%5D+=+-b%2F%282a%29+
+-2+=+-b%2F%282a%29+
+-b+=+-4a+
+b+=+4a+
-----------
So far I have
+y+=+ax%5E2+%2B+4a%2Ax+%2B+c+
Plug in the vertex point ( -2,-8 )
+-8+=+a%2A%28-2%29%5E2+%2B+4a%2A%28-2%29+%2B+c+
+-8+=+4a+-+8a+%2B+c+
+-8+=+-4a+%2B+c+
+c+=+4a+-+8+
Now I have
+y+=+ax%5E2+%2B+4a%2Ax+%2B+4a+-+8+
Plug in ( 0,4 )
+4+=+a%2A0%5E2+%2B+4a%2A0+%2B+4a+-+8+
+4+=+4a+-+8+
+4a+=+12+
+a+=+3+
------------
+y+=+ax%5E2+%2B+4a%2Ax+%2B+4a+-+8+
+y+=+3x%5E2+%2B+4%2A3%2Ax+%2B+4%2A3+-+8+
+y+=+3x%5E2+%2B+12x+%2B+4+
-----------------------
Use the quadratic formula
+x+=+%28-b+%2B-+sqrt%28+b%5E2+-+4%2Aa%2Ac+%29%29+%2F+%282%2Aa%29+
+a+=+3+
+b+=+12+
+c+=+4+
+x+=+%28-12+%2B-+sqrt%28+12%5E2+-+4%2A3%2A4+%29%29+%2F+%282%2A3%29+
+x+=+%28-12+%2B-+sqrt%28+144+-+48+%29%29+%2F+6+
+x+=+%28-12+%2B-+sqrt%28+96+%29%29+%2F+6+
+x+=+%28+-12+%2B+4%2Asqrt%286%29+%29+%2F+6+
+x+=+%28+-6+%2B+2%2Asqrt%286%29+%29+%2F+3+
+x+=+%28+-6+%2B+2%2A2.4495+%29+%2F+3+
+x+=+%28+-6+%2B+4.89898+%29+%2F+3+
+x+=+-1.10102+%2F+3+
+x+=+-.367007+
and
+x+=+%28+-6+-+4.89898+%29+%2F+3+
+x+=+-10.89898+%2F+3+
+x+=+-3.63299+
-------------------
The x-intercepts are at -.367007 and -3.63299
check:
Is the y-intercept at ( 0,4 ) ?
+4+=+3%2A0%5E2+%2B+12%2A0+%2B+4+
+4+=+4+
OK
Does it go through ( -2,-8 ) ?
+y+=+3x%5E2+%2B+12x+%2B+4+
+-8+=+3%28-2%29%5E2+%2B+12%2A%28-2%29+%2B+4+
+-8+=+3%2A4+-+24+%2B+4+
+-8+=+-12+%2B+4+
+-8+=+-8+
OK
Here's the plot:
+graph%28+400%2C+400%2C+-6%2C+3%2C+-10%2C+10%2C+3x%5E2+%2B+12x+%2B+4+%29+